# If $a$ is a sum of squares then there exists an element whose norm is $-1$.

This is part of a question(2.9.3) from Patrick Morandi's "Field and Galois Theory" I am just plain stuck on this fact:

$$F(\sqrt{a})/F$$ is a field extension where $$a\in F-F^2$$ and $$F$$ does not contain a primitive fourth root of unity. If $$a$$ is a sum of two squares in $$F$$, prove that $$N_{F(\sqrt{a})/F}(\alpha)=-1$$ for some $$\alpha\in F(\sqrt{a})$$ and that $$N_{F(\sqrt{a})/F}(\alpha)=a$$ for some $$\alpha\in F(\sqrt{a})$$. Hence, show that $$N_{F(\sqrt{a})/F}(b)\equiv a \mod F^{*2}$$ for some $$b\in F(\sqrt{a})$$.

I know that $$\text{Gal}(F(\sqrt{a})/F=\{1,\sigma\}$$ where $$\sigma:\sqrt{a}\mapsto-\sqrt{a}$$. Also, $$\{1,\sqrt{a}\}$$ is a basis for $$F(\sqrt{a})$$ over $$F$$, so the norm of any element $$c_1+c_2\sqrt{a}\in F(\sqrt{a})$$ is just $$\left(c_1+c_2\sqrt{a}\right)\left(c_1-c_2\sqrt{a}\right)=c_1^2-c_2^2a$$.

The fact that the norm map is surjective when $$F$$ is a finite field gives us the first two facts right away. But it has nothing to do with $$a$$ being a sum of squares, or that $$F$$ has no fourth root of unity, and of course it only works for finite fields.

Also, if you compute $$\min(\sqrt{a-1},F)=x^2-a-1$$, then $$N_{F(\sqrt{a})/F}=(-1)^n(-1)^{\frac{n}{2}}=-1$$ since $$n=[F(\sqrt{a}):F]=2$$. but this also seems wrong because it still has nothing to do with the premises.

How might I proceed here?

• $\sqrt{a-1}$ is not in your field – reuns Aug 10 at 3:51

$$a = u^2+v^2, \qquad N_{F(\sqrt{a})/F}(\frac{u+\sqrt{a}}{v}) =\frac{u+\sqrt{a}}{v}\frac{u-\sqrt{a}}{v}= \frac{u^2-a}{v^2} =-1,\\ \qquad N_{F(\sqrt{a})/F}(\sqrt{a}\frac{u+\sqrt{a}}{v}) = - N_{F(\sqrt{a})/F}(\sqrt{a}) = a$$
Your notations are a bit shaky: $$L$$ is not defined, although it should naturally mean $$F(\sqrt a)$$ ; in the first question, the two $$\alpha$$ should not be the same. This being said, $$a$$ is a sum of squares in $$F$$ iff $$a$$ is a norm from $$F(\sqrt {-1})$$ (which is a quadratic extension of $$F$$ by hypothesis), and the latter condition is plainly equivalent to $$-1$$ being a norm from $$F(\sqrt a)$$. Since $$-a$$ is already a norm from $$F(\sqrt a)$$, so will be $$a$$.